Question

Two concentric coplanar circular loops of radii `r_(1)` and `r_(2)` carry currents of respectively `i_(1)` and `i_(2)` in opposite direction (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to `i_(1)` alone at the centre. if `r_(2)=2r_(1)`. the value of `i_(2)//i_(1)` is
A. `2`
B. `1//2`
C. `1//4`
D. `1`

Answer 1

Correct Answer - D
Magnetic field at centre due to smaller loop
`B_(1)=(mu_(0))/(4pi). (2pii_(1))/(r_(1)).......(i)`
Due to Bigger loop `B_(2)=(mu_(0))/(4pi). (2pii_(2))/(r_(2)).......(i)` So net magnetic field at centre `B=B_(1)-B_(2)=(mu_(0))/(4pi)xx2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))`
According to question `B=1/2xxB_(1)`
`(mu_(0))/(2pi).2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))=1/2xx(mu_(0))/(4pi). (2pii_(1))/(r_(1))`
`(i_(1))/(r_(1))-(i_(2))/(r_(2))=(i_(1))/(2r_(1)) implies (i_(1))/(2r_(1))=(i_(2))/(r_(2)) implies (i_(2))/(r_(2))=1{r_(2)=2r_(1)}`